| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Curve with minimum point |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard techniques: (i) differentiate using product rule to find the minimum (dy/dx = 2x ln x + x = 0 gives x = e^{-1/2}), and (ii) integrate by parts twice to find the area. Both are routine applications of A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for \(x\) | M1 | |
| Obtain answer \(x = e^{-\frac{2}{3}}\), or equivalent | A1 | |
| Obtain answer \(y = -\frac{1}{2} e^{-1}\), or equivalent | A1 | [5] |
| (ii) Attempt integration by parts reaching \(kx^3 \ln x \pm k \int x^3 \cdot \frac{1}{x} dx\) | M1* | |
| Obtain \(\frac{1}{3} x^3 \ln x - \frac{1}{3} \int x^2 dx\), or equivalent | A1 | |
| Integrate again and obtain \(\frac{1}{3} x^3 \ln x - \frac{1}{9} x^3\), or equivalent | A1 | |
| Use limits \(x = 1\) and \(x = e\), having integrated twice | M1(dep*) | |
| Obtain answer \(\frac{1}{9} (2e^3 + 1)\), or exact equivalent | A1 | [5] |
| [SR: An attempt reaching \(ax^2 (x \ln x - x) + b \int 2x(x \ln x - x) dx\) scores M1. Then give the first A1 for \(I = x^2 (x \ln x - x) - 2I + \int 2x^2 dx\), or equivalently.] |
**(i)** Use product rule | M1 |
Obtain correct derivative in any form | A1 |
Equate derivative to zero and solve for $x$ | M1 |
Obtain answer $x = e^{-\frac{2}{3}}$, or equivalent | A1 |
Obtain answer $y = -\frac{1}{2} e^{-1}$, or equivalent | A1 | [5]
**(ii)** Attempt integration by parts reaching $kx^3 \ln x \pm k \int x^3 \cdot \frac{1}{x} dx$ | M1* |
Obtain $\frac{1}{3} x^3 \ln x - \frac{1}{3} \int x^2 dx$, or equivalent | A1 |
Integrate again and obtain $\frac{1}{3} x^3 \ln x - \frac{1}{9} x^3$, or equivalent | A1 |
Use limits $x = 1$ and $x = e$, having integrated twice | M1(dep*) |
Obtain answer $\frac{1}{9} (2e^3 + 1)$, or exact equivalent | A1 | [5]
[SR: An attempt reaching $ax^2 (x \ln x - x) + b \int 2x(x \ln x - x) dx$ scores M1. Then give the first A1 for $I = x^2 (x \ln x - x) - 2I + \int 2x^2 dx$, or equivalently.] | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{9e129863-5994-4e17-81f8-e139515998d1-3_666_956_1231_593}
The diagram shows the curve $y = x ^ { 2 } \ln x$ and its minimum point $M$.\\
(i) Find the exact values of the coordinates of $M$.\\
(ii) Find the exact value of the area of the shaded region bounded by the curve, the $x$-axis and the line $x = \mathrm { e }$.
\hfill \mbox{\textit{CAIE P3 2011 Q9 [10]}}